Solve for d R/(p(d+a))=g

$\frac{R}{p (d + a)} = g$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$p (d + a), 1$

The LCM of one and any expression is the expression.

$p (d + a)$

Multiply each term in $\frac{R}{p (d + a)} = g$ by $p (d + a)$ to eliminate the fractions.

Multiply each term in $\frac{R}{p (d + a)} = g$ by $p (d + a)$ .

$\frac{R}{p (d + a)} (p (d + a)) = g (p (d + a))$

Simplify the left side.

Cancel the common factor of $p (d + a)$ .

Cancel the common factor.

$\frac{R}{p (d + a)} (p (d + a)) = g (p (d + a))$

Rewrite the expression.

$R = g (p (d + a))$

Simplify the right side.

Apply the distributive property.

$R = g (p d + p a)$

Apply the distributive property.

$R = g p d + g p a$

Solve the equation.

Rewrite the equation as $g p d + g p a = R$ .

$g p d + g p a = R$

Subtract $g p a$ from both sides of the equation.

$g p d = R - g p a$

Divide each term in $g p d = R - g p a$ by $g p$ and simplify.

Divide each term in $g p d = R - g p a$ by $g p$ .

$\frac{g p d}{g p} = \frac{R}{g p} + \frac{- g p a}{g p}$

Simplify the left side.

Cancel the common factor of $g$ .

Cancel the common factor.

$\frac{g p d}{g p} = \frac{R}{g p} + \frac{- g p a}{g p}$

Rewrite the expression.

$\frac{p d}{p} = \frac{R}{g p} + \frac{- g p a}{g p}$

Cancel the common factor of $p$ .

Cancel the common factor.

$\frac{p d}{p} = \frac{R}{g p} + \frac{- g p a}{g p}$

Divide $d$ by $1$ .

$d = \frac{R}{g p} + \frac{- g p a}{g p}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $g$ .

Cancel the common factor.

$d = \frac{R}{g p} + \frac{- g p a}{g p}$

Rewrite the expression.

$d = \frac{R}{g p} + \frac{- 1 p a}{p}$

Cancel the common factor of $p$ .

Cancel the common factor.

$d = \frac{R}{g p} + \frac{- 1 p a}{p}$

Divide $- 1 a$ by $1$ .

$d = \frac{R}{g p} - 1 a$

Rewrite $- 1 a$ as $- a$ .

$d = \frac{R}{g p} - a$

Solve for d R/(p(d+a))=g

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